HSC Exam Practice

Sample response. The cyclotron frequency is the number of complete circular orbits per second that a charged particle completes in a uniform magnetic field: \(f = qB / 2\pi m\). It depends only on the charge-to-mass ratio \(q/m\) of the particle and the magnetic field strength \(B\). It is independent of the particle’s speed.

Marking criteria. 1 mark for defining cyclotron frequency as revolutions per second (or \(f = 1/T\)). 1 mark for stating the formula \(f = qB/2\pi m\) correctly. 1 mark for identifying that it depends on \(q/m\) and \(B\) only (not speed).

Sample response. When a charged particle moves perpendicular to a magnetic field, it experiences a magnetic force \(F = qvB\) directed toward the centre of the circular path. This force acts as the centripetal force required for circular motion: \(F_c = mv^2/r\). Setting them equal: \(qvB = mv^2/r\). Dividing both sides by \(v\): \(qB = mv/r\). Rearranging: \(r = mv/qB\).

Marking criteria. 1 mark for correctly identifying \(F_{mag} = qvB\) and \(F_c = mv^2/r\) as the forces being equated. 1 mark for correct algebraic setup \(qvB = mv^2/r\). 1 mark for correct rearrangement to \(r = mv/qB\) with clear working shown.

Sample response. The magnetic force on a charged particle is always perpendicular to its velocity. Work is defined as \(W = Fd\cos\theta\); when force and displacement are perpendicular, \(\theta = 90°\) and \(\cos 90° = 0\), so no work is done. By the work-energy theorem, if no work is done, the kinetic energy (and therefore speed) cannot change. However, because the force is always directed toward the centre of the circular path, it continuously changes the direction of the velocity without changing its magnitude, producing uniform circular motion.

Marking criteria. 1 mark for stating the magnetic force is perpendicular to velocity. 1 mark for applying \(W = Fd\cos 90° = 0\) (or equivalent: no work done means no energy transfer). 1 mark for explaining that force perpendicular to velocity changes direction but not speed.

Sample response. The orbital radius \(r = mv/qB\) is directly proportional to speed (\(r \propto v\)): doubling the speed doubles the radius, so faster particles travel in larger circles. By contrast, the orbital period \(T = 2\pi m/qB\) contains no \(v\) term at all. The period is completely independent of speed: a faster particle travels its larger circle in exactly the same time as a slower particle travels its smaller circle. This remarkable property arises because both the circumference and the speed increase by the same factor, leaving the time unchanged.

Marking criteria. 1 mark for stating \(r \propto v\) with formula \(r = mv/qB\). 1 mark for the physical consequence (faster particle, larger circle). 1 mark for stating \(T = 2\pi m/qB\) contains no \(v\). 1 mark for explaining why T is independent (circumference and speed scale identically).

(a) \(r_p = m_p v / (eB) = (1.67\times10^{-27})(3.5\times10^6) / [(1.60\times10^{-19})(0.025)] = 5.845\times10^{-21} / 4.00\times10^{-21} = \mathbf{1.46\ \text{m}}\). [1 mark for correct substitution and answer with unit]

(b) \(T_p = 2\pi m_p / (eB) = 2\pi(1.67\times10^{-27}) / [(1.60\times10^{-19})(0.025)] = 1.050\times10^{-26} / 4.00\times10^{-21} = \mathbf{2.62\times10^{-6}\ \text{s}}\). [1 mark]

(c) Electron radius: much smaller than the proton’s. From \(r = mv/qB\), with the same \(v\), \(q\), and \(B\), \(r \propto m\). Since \(m_e \approx m_p/1836\), the electron’s radius is \(\approx 1836\) times smaller: \(r_e \approx 1.46/1836 \approx 7.95\times10^{-4}\ \text{m}\) [1 mark]. Electron period: much shorter than the proton’s. From \(T = 2\pi m/qB\), \(T \propto m\), so the electron’s period is \(\approx 1836\) times shorter: \(T_e \approx 2.62\times10^{-6}/1836 \approx 1.43\times10^{-9}\ \text{s}\) [1 mark].

Sample response. In a cyclotron, charged particles orbit in two D-shaped electrodes within a magnetic field and are accelerated each time they cross the gap between the D’s by an alternating electric field. As particles gain energy, their orbital radius increases, but their orbital period \(T = 2\pi m/qB\) remains constant because it does not depend on speed. The alternating voltage can therefore be fixed at the cyclotron frequency \(f = qB/2\pi m\): it will always be in the correct phase to accelerate the particle across the gap at every half-revolution, regardless of how fast the particle is moving. This is what makes the cyclotron self-consistent: the frequency of the accelerating voltage is set once and never needs adjusting as particles gain energy.

Marking criteria. 1 mark for explaining the alternating electric field accelerates particles each half-revolution. 1 mark for stating that the period is independent of speed (\(T = 2\pi m/qB\), no \(v\)). 1 mark for linking this to the fixed frequency of the alternating voltage staying in phase as particles gain speed.

Completed table (B = 0.080 T, v = 4.0 × 10⁶ m/s):

Electron: \(r_e = (9.11\times10^{-31})(4.0\times10^6)/[(1.60\times10^{-19})(0.080)] = 3.644\times10^{-24}/1.28\times10^{-20} = \mathbf{2.85\times10^{-4}\ \text{m}}\). \(T_e = 2\pi(9.11\times10^{-31})/(1.60\times10^{-19}\times0.080) = \mathbf{4.47\times10^{-10}\ \text{s}}\).

Proton: \(r_p = (1.67\times10^{-27})(4.0\times10^6)/(1.28\times10^{-20}) = \mathbf{0.522\ \text{m}}\). \(T_p = 2\pi(1.67\times10^{-27})/(1.28\times10^{-20}) = \mathbf{8.20\times10^{-7}\ \text{s}}\).

Alpha: \(r_\alpha = (6.64\times10^{-27})(4.0\times10^6)/[(3.20\times10^{-19})(0.080)] = 2.656\times10^{-20}/2.56\times10^{-20} = \mathbf{1.04\ \text{m}}\). \(T_\alpha = 2\pi(6.64\times10^{-27})/(3.20\times10^{-19}\times0.080) = \mathbf{1.63\times10^{-6}\ \text{s}}\).

Marking criteria for (a): 1 mark for full working for electron radius; 1 mark for correct electron radius; 1 mark for correct electron period; 1 mark for correct proton radius; 1 mark for correct alpha radius. (Periods for proton and alpha may be calculated or inferred by ratio; 1 mark per correct value to 3 sig. figs.)

Sample response for (b): \(T_\alpha / T_p = (m_\alpha/q_\alpha)/(m_p/q_p) = (6.64\times10^{-27}/3.20\times10^{-19})/(1.67\times10^{-27}/1.60\times10^{-19}) = (2.075\times10^{-8})/(1.044\times10^{-8}) = \mathbf{1.99 \approx 2.0}\). The alpha’s period is approximately twice the proton’s. This occurs because \(T \propto m/q\): the alpha’s mass is about 4 times the proton’s but its charge is 2 times greater, giving a mass-to-charge ratio (\(m/q\)) that is twice the proton’s. A larger \(m/q\) means the particle is harder to deflect per unit charge, so it takes longer to complete one orbit.

Marking criteria for (b): 1 mark for correct numerical ratio (approx 2.0). 1 mark for citing \(T \propto m/q\) and comparing the \(m/q\) ratios. 1 mark for a clear physical explanation linking the larger \(m/q\) to a longer period.

Sample Band 6 response. Both a charged particle in a magnetic field and a satellite in orbit undergo uniform circular motion — their speeds are constant and the centripetal force is always directed toward the centre of the circular path. The fundamental method of analysis is identical: identify the real force, set it equal to the required centripetal force \(mv^2/r\), and solve for the radius.

For a charged particle, the centripetal force is provided by the magnetic force: \(F = qvB\). Setting \(qvB = mv^2/r\) and solving gives \(r = mv/qB\). This shows that \(r \propto v\): a faster particle orbits in a proportionally larger circle. For a satellite, the centripetal force is gravitational: \(F = GMm/r^2\). Setting \(GMm/r^2 = mv^2/r\) and solving gives \(r = GM/v^2\), showing \(r \propto 1/v^2\): a faster satellite must orbit at a smaller radius (lower altitude). This is the key difference in how radius scales with speed — they are opposite.

In both systems, the centripetal force is always perpendicular to the velocity. Work is defined as \(W = F \cdot d \cdot \cos\theta\); with \(\theta = 90°\) in both cases, no work is done by the centripetal force. By the work-energy theorem, zero net work means no change in kinetic energy and therefore constant speed. This is why both a satellite and a charged particle maintain constant speed in their orbits without any engine or accelerating force.

The key difference that makes a cyclotron useful for particle physics is the independence of orbital period from speed (\(T = 2\pi m/qB\), no \(v\)). As protons are repeatedly accelerated, their period stays constant, keeping them in phase with the fixed-frequency alternating voltage. This allows particles to be accelerated to very high energies within a compact device. A magnetic field cannot maintain a satellite orbit because gravity, not magnetism, acts on a satellite; and even if a magnetic force could act, the \(r = mv/qB\) scaling (larger orbit at higher speed) would mean that an increase in energy would move the orbit outward, not downward, making the trajectory unstable in a way that is not useful for orbital mechanics.

Marking criteria (7 marks): 1 = correctly identifies and writes the centripetal-force equation for each system (\(qvB = mv^2/r\) and \(GMm/r^2 = mv^2/r\)). 1 = derives both radius formulas and states the correct proportionality (\(r \propto v\) for particle; \(r \propto 1/v^2\) for satellite). 1 = explicitly notes the key difference (radius increases with speed for particle, decreases for satellite). 1 = explains that no work is done because force is perpendicular to velocity (cites \(W = Fd\cos 90° = 0\) or equivalent). 1 = links zero work to constant speed via the work-energy theorem (applies to both systems). 1 = clearly explains how the period-independence enables a cyclotron (fixed frequency stays in phase as particles gain energy). 1 = reaches an explicit evaluative judgement about why this property is unique to the charged-particle/magnetic system and not applicable to satellite orbits, integrating at least two of the points above.